# Thread: Another Trig Problem

1. ## Another Trig Problem

If your not too busy...

1-tan^2X =1-2tan^2X
1+tan^2X

2. Originally Posted by Dominic O'L
If your not too busy...

1-tan^2X =1-2tan^2X
1+tan^2X
What is the question? Solve for x ....?

3. No, just prove that they are equal.

4. Originally Posted by Dominic O'L
No, just prove that they are equal.
They're not. eg. Does it work for $x = \frac{\pi}{4}$ ....?

5. oh sorry i read the question incorrectly

1-tan^2X =1-2sin^2X
1+tan^2X

6. Originally Posted by Dominic O'L
oh sorry i read the question incorrectly

1-tan^2X =1-2sin^2X
1+tan^2X
First, use long division on the left hand side.

You should find

$\frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\tan^2{x} + 1}$

Remember that $\tan^2{x} + 1 = \sec^2{x}$

So $\frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\sec^2{x}}$

$= -1 + 2\cos^2{x}$

$= -1 + 2(1 - \sin^2{x})$, since $\cos^2{x} + \sin^2{x} = 1$...

$= -1 + 2 - 2\sin^2{x}$

$= 1 - 2\sin^2{x}$.

7. Sorrry, but I think I'm supposed to solve it without sec.

8. Originally Posted by Dominic O'L
Sorrry, but I think I'm supposed to solve it without sec.
Then write it as

$-1 + \frac{2}{\frac{1}{\cos^2{x}}}$.

Same thing.

9. Hello Dominc O'L

$\frac{1-\tan^2x}{1+\tan^2x}$

$=\frac{1-\tan^2x}{1+\tan^2x}\cdot \frac{\cos^2x}{\cos^2x}$

$=\frac{\cos^2x - \sin^2x}{\cos^2x + \sin^2x}$

$= \frac{(1-\sin^2x)-\sin^2x}{1}$

$= 1 - 2\sin^2x$